Elementary Geometry for College Students (7th Edition) Clone

Published by Cengage
ISBN 10: 978-1-337-61408-5
ISBN 13: 978-1-33761-408-5

Appendix A - A.4 - Quadratic Equations - Exercises - Page 556: 29

Answer

$x=-\frac{2}{3}~~$ or $~~x=4$

Work Step by Step

Let's consider a trinomial in this form: $~ax^2+bx+c$ To factor this trinomial, we need to find two numbers $r$ and $s$ such that $r+s = b$ and $r\times s = a\times c$ Then the next step is to write the trinomial as follows: $~x^2+bx+c = ax^2+rx+sx+c$ We can rewrite the given equation: $3x^2 = 10x+8$ $3x^2 - 10x - 8 = 0$ To factor the left side of this equation, we need to find two numbers $r$ and $s$ such that $r+s = -10~$ and $~r\times s = -24$. We can see that $(-12)+(2) = -10~$ and $(-12)\times (2) = -24$ We can solve the equation as follows: $3x^2 = 10x+8$ $3x^2 - 10x - 8 = 0$ $3x^2 - 12x+2x - 8 = 0$ $(3x^2 - 12x)+(2x - 8) = 0$ $(3x)(x - 4)+(2)(x - 4) = 0$ $(3x+2)~(x-4) = 0$ $3x+2 = 0~~$ or $~~x-4 = 0$ $x=-\frac{2}{3}~~$ or $~~x=4$
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